Fundamentals of physics 9e halliday resnick walker solutions manual and test bank
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Book Description
Publication Date: March 16, 2010  ISBN10: 0470469110  ISBN13: 9780470469118  Edition: 9This book arms engineers with the tools to apply key physics concepts in the field. A number of the key figures in the new edition are revised to provide a more inviting and informative treatment. The figures are broken into component parts with supporting commentary so that they can more readily see the key ideas. Material from The Flying Circus is incorporated into the chapter opener puzzlers, sample problems, examples and endofchapter problems to make the subject more engaging. Checkpoints enable them to check their understanding of a question with some reasoning based on the narrative or sample problem they just read. Sample Problems also demonstrate how engineers can solve problems with reasoned solutions.
INCLUDES PARTS 14
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Import Settings:
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Information Field: Difficulty
Information Field: Section
Highest Answer Letter: E
Multiple Keywords in Same
Paragraph: No
Chapter: Chapter 2
Multiple Choice
1. A particle moves along the x axis from
x_{i} to x_{ f }. Of the
following values of the initial and final coordinates, which results in the
displacement with the largest magnitude?
A) x_{i} = 4m, x_{ f }=
6m
B) x_{i} = –4m, x_{ f }=
–8m
C) x_{i}_{ }= –4m, x_{
f }= 2m
D) x_{i} = 4m, x_{ f }=
–2m
E) x_{i} = –4m, x_{ f }=
4m
Ans: E
Difficulty: E
Section: 23
2. A particle moves along the x axis from
x_{i} to x_{ f }.
Of the following values of the initial and final coordiantes, which
results in a negative displacement?
A) x_{i} = 4m, x_{ f }=
6m
B) x_{i} = –4m, x_{ f }=
–8m
C) x_{i} = –4m, x_{ f }=
2m
D) x_{i} = –4m, x_{ f }=
–2m
E) x_{i} = –4m, x_{ f }= 4m
Ans: B
Difficulty: E
Section: 23
3. The position y of a particle moving
along the y axis depends on the time t according to the equation
y = at – bt^{2}. The dimensions of
the quantities a and b are respectively:
A) L^{2}/T, L^{3}/T^{2}
B) L/T^{2}, L^{2}/T
C) L/T, L/T^{2}
D) L^{3}/T, T^{2}/L
E) none of these
Ans: C
Difficulty: E
Section: 23
4. The average speed of a moving object during a
given interval of time is always:
A) the magnitude of its average velocity over
the interval
B) the distance covered during the time interval
divided by the time interval
C) onehalf its speed at the end of the interval
D) its acceleration multiplied by the time
interval
E) onehalf its acceleration multiplied by the
time interval.
Ans: B
Difficulty: E
Section: 24
5. Two automobiles are 150 kilometers apart and
traveling toward each other. One automobile is moving at 60 km/h and the other
is moving at 40 km/h. In how many hours will they meet?
A) 2.5
B) 2.0
C) 1.75
D) 1.5
E) 1.25
Ans: D
Difficulty: E
Section: 24
6. A car travels 40 kilometers at an average
speed of 80 km/h and then travels 40 kilometers at an average speed of 40 km/h.
The average speed of the car for this 80 km trip is:
A) 40 km/h
B) 45 km/h
C) 48 km/h
D) 53 km/h
E) 80 km/h
Ans: D
Difficulty: M
Section: 24
7. A car starts from Hither, goes 50 km in a
straight line to Yon, immediately turns around, and returns to Hither. The time
for this round trip is 2 hours. The magnitude of the average velocity of the
car for this round trip is:
A) 0
B) 50 km/hr
C) 100 km/hr
D) 200 km/hr
E) cannot be calculated without knowing the
acceleration
Ans: A
Difficulty: E
Section: 24
8. A car starts from Hither, goes 50 km in a
straight line to Yon, immediately turns around, and returns to Hither. The time
for this round trip is 2 hours. The average speed of the car for this round
trip is:
A) 0
B) 50 km/h
C) 100 km/h
D) 200 km/h
E) cannot be calculated without knowing the
acceleration
Ans: B
Difficulty: E
Section: 24
9. The coordinate of an object is given as a
function of time by x = 7t – 3t^{2}, where x
is in meters and t is in seconds.
Its average velocity over the interval from t = 0 to t = 2
s is:
A) 5 m/s
B) –5 m/s
C) 11 m/s
D) –11 m/s
E) –14.5 m/s
Ans: B
Difficulty: M
Section: 24
10. The coordinate of a particle in meters is
given by x(t) = 16t – 3.0t^{3}, where the
time t is in seconds. The particle is momentarily at rest at t =
A) 0.75 s
B) 1.3 s
C) 5.3 s
D) 7.3 s
E) 9.3 s
Ans: B
Difficulty: M
Section: 25
Chapter 6
1. The greatest deceleration (of magnitude a) is provided by the maximum friction
force (Eq. 61, with F_{N}
= mg in this case). Using Newton ’s
second law, we find
a = f_{s,max }/m = m_{s}g.
Eq. 216 then gives the shortest distance
to stop: Dx = v^{2}/2a = 36 m .
In this calculation, it is important to first convert v to 13 m /s.
2. Applying Newton ’s second law to the horizontal motion,
we have F  m_{k }m_{ }g = ma, where we have used Eq.
62, assuming that F_{N}
= mg (which is equivalent to assuming
that the vertical force from the broom is negligible). Eq. 216 relates the
distance traveled and the final speed to the acceleration: v^{2 }=
2a Dx. This gives a = 1.4 m /s^{2}.
Returning to the force equation, we find (with F = 25 N and m = 3.5 kg ) that m_{k }= 0.58.
3. The freebody diagram for the bureau
is shown to the right. We do not consider the possibility that the bureau
might tip, and treat this as a purely horizontal motion problem (with the
person’s push in the +x direction). Applying
respectively. The second equation yields
the normal force F_{N}_{
}= mg, whereupon the maximum
static friction is found to be (from Eq. 61) .


Thus, the first equation becomes
where we have set a = 0 to be consistent with the fact that the static friction is
still (just barely) able to prevent the bureau from moving.
(a) With and m = 45 kg ,
the equation above leads to F = 198
N.
To bring the bureau into a state of motion,
the person should push with any force greater than this value. Rounding to two
significant figures, we can therefore say the minimum required push is F = 2.0 ´ 10^{2} N.
(b) Replacing m = 45 kg
with m = 28 kg , the reasoning above leads to roughly .
Note: The values found above represent the
minimum force required to overcome the friction. Applying a force greater than results in a net force
in the +xdirection, and hence, nonzero acceleration.
4. We first analyze the forces on the pig
of mass m. The incline angle is q.
The +x
direction is “downhill.’’ Application of Newton ’s
second law to the x and yaxes leads to
Solving these along with Eq. 62 (f_{k} = m_{k}F_{N}) produces the following result for the pig’s downhill acceleration:
To compute the time to slide from rest
through a downhill distance , we use Eq. 215:
We denote the frictionless (m_{k} = 0) case with a prime and set
up a ratio:
which leads us to conclude that if t/t' = 2 then a' = 4a . Putting in what we
found out above about the accelerations, we have
Using q = 35°, we obtain m_{k} = 0.53.
5. In addition to the forces already shown
in Fig. 617, a freebody diagram would include an upward normal force exerted by the floor
on the block, a downward representing the gravitational
pull exerted by Earth, and an assumedleftward for the kinetic or
static friction. We choose +x
rightwards and +y upwards. We apply Newton ’s second law to
these axes:
where F = 6.0 N and m = 2.5
kg is the mass of the block.
(a) In this case, P = 8.0 N leads to
F_{N} = (2.5 kg )(9.8 m /s^{2}) – 8.0 N = 16.5 N.
Using Eq. 61, this implies , which is larger than the 6.0 N rightward force – so the
block (which was initially at rest) does not move. Putting a = 0 into the first of our equations above yields a static
friction force of f = P = 6.0 N.
(b) In this case, P = 10 N, the normal force is
F_{N} = (2.5 kg )(9.8 m /s^{2}) – 10 N = 14.5 N.
Using Eq. 61, this implies , which is less than the 6.0 N rightward force – so the block
does move. Hence, we are dealing not with static but with kinetic friction,
which Eq. 62 reveals to be .
(c) In this last case, P = 12 N leads to F_{N}
= 12.5 N and thus to , which (as expected) is less than the 6.0 N rightward force
– so the block moves. The kinetic friction force, then, is .
6. The
freebody diagram for the player is shown to the right. is the normal force of
the ground on the player, is the force of
gravity, and is the force of
friction. The force of friction is related to the normal force by f = m_{k}F_{N}. We use Newton ’s
second law applied to the vertical axis to find the normal force. The vertical
component of the acceleration is zero, so we obtain F_{N} – mg = 0; thus, F_{N}
= mg. Consequently,
7. The freebody diagram for the crate is
shown to the right. We denote as the horizontal
force of the person exerted on the crate (in the +x direction), is the force of
kinetic friction (in the –x
direction), is the vertical
normal force exerted by the floor (in the +y direction), and is the force of
gravity. The magnitude of the force of friction is given by (Eq. 62):
f_{k} = m_{k}F_{N} .


Applying Newton ’s second law to the x and y axes, we obtain
respectively.
(a) The second equation above yields the
normal force F_{N}_{
}= mg, so that the friction is
(b) The first equation becomes
which (with F = 220 N) we solve to find
Note: For the crate to accelerate, the
condition must be met. As can be
seen from the equation above, the greater the value of , the smaller the acceleration with the same applied force.
8. To maintain the stone’s motion, a
horizontal force (in the +x
direction) is needed that cancels the retarding effect due to kinetic friction.
Applying Newton ’s
second to the x and y axes, we obtain
respectively. The second equation yields
the normal force F_{N}
= mg, so that (using Eq. 62) the
kinetic friction becomes f_{k}
= m_{k} mg. Thus, the first
equation becomes
where we have set a = 0 to be consistent with the idea that the horizontal velocity
of the stone should remain constant. With m
= 20 kg and m_{k} = 0.80, we find F = 1.6 ´ 10^{2} N.
9. We choose +x horizontally rightwards and +y
upwards and observe that the 15 N force has components F_{x} = F cos q and F_{y} = – F sin q.
(a) We apply Newton ’s second law to the y axis:
With m_{k} = 0.25, Eq. 62 leads to f_{k} = 11 N.
(b) We apply Newton ’s second law to the x axis:
.
Since the result is positivevalued, then
the block is accelerating in the +x
(rightward) direction.
10. (a) The freebody diagram for the block
is shown below, with being the force
applied to the block, the normal force of
the floor on the block, the force of gravity,
and the force of friction.
We take the +x direction to be
horizontal to the right and the +y
direction to be up. The equations for the x
and the y components of the force according to Newton ’s second law are:


Now f
=m_{k}F_{N}, and the second equation gives F_{N} = mg – Fsinq, which yields . This expression is substituted for f in the first equation to obtain
F cos q – m_{k} (mg
– F sin q) = ma,
so the acceleration is
.
(a) If and then the magnitude of has a maximum value of
On the other hand, Therefore, and the block remains
stationary with .
(b) If and then the magnitude of has a maximum value of
In this case, Therefore, the
acceleration of the block is
11. (a)
The freebody diagram for the crate is shown on the right. is the tension force
of the rope on the crate, is the normal force of
the floor on the crate, is the force of
gravity, and is the force of
friction. We take the +x direction to
be horizontal to the right and the +y
direction to be up. We assume the crate is motionless. The equations for the x and the y components of the
force according to Newton ’s
second law are:
T cos q – f = 0
where q = 15° is the angle between the rope and the horizontal. The first
equation gives f = T cos q and the second gives F_{N}
= mg – T sin q. If the
crate is to remain at rest, f must be
less than m_{s} F_{N}, or T cos q < m_{s} (mg – T sinq). When the tension force is sufficient to just start the crate
moving, we must have
T cos q = m_{s} (mg
– T sin q).
We solve for the tension:
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